A brass cube of side a and density `sigma`is floating in mercury of density `rho`. If the cube is displaced a bit vertically, it executes S.H.M. Its time period will be

Options

A. `2pisqrt((sigmaa)/(rhog))`
B. `2pisqrt((rhoa)/(sigmag))`
C. `2pisqrt((rhog)/(sigmaa))`
D. `2pisqrt((sigmag)/(rhoa))`

Answer

To solve the problem of finding the time period of a brass cube floating in mercury when it is displaced slightly, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Parameters:**
- Side of the brass cube = \( a \)
- Density of brass = \( \sigma \)
- Density of mercury = \( \rho \)

2. **Understand the Displacement:**
- When the cube is displaced vertically by a small distance \( \delta x \), it will experience a change in the buoyant force.

3. **Calculate the Buoyant Force:**
- The volume of the cube submerged in mercury when it is at equilibrium is equal to the volume of the cube, which is \( a^3 \).
- The change in buoyant force \( \Delta B \) when the cube is displaced by \( \delta x \) is given by:
\[ \Delta B = \text{Area} \times \text{Height} \times \text{Density} \times g = a^2 \cdot \delta x \cdot \rho \cdot g \]
- Thus, the change in buoyant force is:
\[ \Delta B = a^2 \delta x \rho g \]

4. **Determine the Restoring Force:**
- The restoring force \( F_R \) acting on the cube is equal to the change in buoyant force:
\[ F_R = -\Delta B = -a^2 \delta x \rho g \]

5. **Relate the Restoring Force to SHM:**
- According to Hooke's law for simple harmonic motion (SHM), the restoring force can be expressed as:
\[ F_R = -k \delta x \]
- Comparing both expressions gives:
\[ k = a^2 \rho g \]

6. **Calculate the Mass of the Cube:**
- The mass \( m \) of the brass cube is:
\[ m = \text{Density} \times \text{Volume} = \sigma \cdot a^3 \]

7. **Use the Formula for Time Period:**
- The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- Substituting the values of \( m \) and \( k \):
\[ T = 2\pi \sqrt{\frac{\sigma a^3}{a^2 \rho g}} = 2\pi \sqrt{\frac{\sigma a}{\rho g}} \]

8. **Final Result:**
- Therefore, the time period \( T \) of the brass cube executing SHM is:
\[ T = 2\pi \sqrt{\frac{\sigma a}{\rho g}} \]